Originally Posted by
NickLappos
So if you restore the power while descending at 250 fpm to 80%, the aircraft continues to have a power deficit and will increase its descent rate. To arrest a descent of 250fpm , you must pull 85% torque. (all these numbers are hypothetical, but illustrate the issue.
Interesting,
very important to memorize, but why is that so?
If reapplying the (ficticious) original HOGE power of 80% after inducing a 250fpm descent does NOT arrest descent rate, e.g. sink rate keeps increasing, this would mean that a descending rotor produces less lift than a hovering rotor (pitch being identical).
I can imagine that as pitch were identical, AoA is higher with the rotor descending.
But this condition should produce MORE lift than in HOGE, right?
I'd say this: if (blade) pitch during -250fpm is kept identical to HOGE pitch, AoA will be higher due to descent. Higher AoA means more drag. More drag at same power setting results in rrpm decay. Rrpm decay is compensated by closed loop rrpm control via increase in torque. Now the increased drag from increased AoA is nicely compensated, pitch and rrpm still at HOGE setting, e.g. same lift as in HOGE, but TQ has climbed to a higher level to reestablish equilibrium. How does that sound?
If above is true, merely wanting to control/arrest a desired descent rate will produce requirement for more torque. I think I got that, not?
I can see now, that a piston powered helo (the ones I'm flying) can effectively run out of power and settle_with_power from trying to achieve too high a sink rate on approach.
But wouldn't that mean, that people descending with insufficient turbine TQ margin would rather overtorque their engine,
(fuel control trying to maintain rrpm), till the gearbox fails, than "falling through" from settling-with-(lackof)-power?
I always picture turbine helos as excessively overpowered with the gearbox as the limiting factor, in non hot-and-high conditions.